how to make a function that calculate consective difference of elements of a vector

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Muhammad Usman Saleem
编辑： Andrei Bobrov 2015-5-27
Hi every one. I am going to attempt that query.. Write a function called neighbor that takes as input a row vector called v and creates another row vector as output that contains the absolute values of the differences between neighboring elements of v. For example, if v == [1 2 4 7], then the output of the function would be [1 2 3]. Notice that the length of the output vector is one less than that of the input. Check that the input v is indeed a vector and has at least two elements and return an empty array otherwise. You are not allowed to use the diff built‐in function. I am trying that code, getting an error in for loop because its inner statement (vi+1) creating problem.To me instead of for loop while loop will be implement. But how i not know.Any correction will be highly appreciable..
function [A]=neighbor(v)
if isvector(v) && length(v)>=2 % checking whether v is a vector and have two elements
for i=1:length(v)
A=v(i+1)-v(i); % getting the absolute values of v(i+1)-v(i)
end
else
A=[];
end
end

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采纳的回答

Thorsten
Thorsten 2015-5-27
编辑：Thorsten 2015-5-27
Why not using your original solution with the changes that Yoav suggested, plus an additional abs() to get the absolute difference?
function [A]=neighbor(v)
if isvector(v) && length(v)>=2 % checking whether v is a vector and have two elements
for i=1:length(v) - 1 % *** -1 otherwise i+1 becomes invalid
A(i)=abs(v(i+1)-v(i)); % getting the absolute values of v(i+1)-v(i)
end
else
A=[];
end
end
Instead of the for loop, you can use
A = abs(v(2:end) - v(1:end-1));
If you run into problems, please report input and the error it produces.
  1 个评论
Muhammad Usman Saleem
Muhammad Usman Saleem 2015-5-27
@ Thorsten thanks.. Actually he write it very briefly.. that why i was not using this.. Your other code has already share by @Stephen Cobeldick .. please see his answer..

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更多回答（1 个）

Yoav Livneh
Yoav Livneh 2015-5-27
If you insisnt on using for loops you just need to change the limit of the for loop to
length(v)-1
and of course add
A(i) = ...
so you get a vector and not just a single value.
But you can just use indexing instead of loops to solve this:
A = v(2:end) - v(1:end-1);

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